To solve the problem, we first assume the standard scenario involving candies for A (x) and B (y) with the following conditions:

Step 1: Set up equations based on the conditions

Let’s assume the conditions are:

1. If A gives 5 candies to B, B has twice as many as A:
   ( y + 5 = 2(x - 5) )
   Simplify: ( y + 5 = 2x - 10 ) → ( 2x - y = 15 ) ...(1)

2. If B gives 5 candies to A, A has thrice as many as B:
   ( x + 5 = 3(y - 5) )
   Simplify: ( x + 5 = 3y - 15 ) → ( x - 3y = -20 ) ...(2)

Step 2: Solve the system of equations

From equation (2): ( x = 3y - 20 )

Substitute ( x = 3y - 20 ) into equation (1):
( 2(3y - 20) - y = 15 )
( 6y - 40 - y = 15 )
( 5y = 55 ) → ( y = 11 )

Substitute ( y = 11 ) into ( x = 3y - 20 ):
( x = 3(11) - 20 = 33 - 20 = 13 )

Verification

• A gives 5 → A=8, B=16 → 16=2×8 (correct).
• B gives5 → A=18, B=6 →18=3×6 (correct).

Answer: ( x = 13 ), ( y = 11 )

(\boxed{x=13, y=11})

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